Answer:
The final temperature of the element = 262.67°C
The power dissipated in the heating element initially = 163.21 W
The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W
Explanation:
Our given parameters include;
A Nichrome heating element operates on 120 V.
Voltage (V) = 120V
Initial Current (I₁) = 1.36 A
Initial Temperature (T₁) = 28°C
Final Current (I₂) = 1.23 A
Final Temperature (T₂) = unknown ????
Temperature dependencies of resistance is given by:
[tex]R_{T(2)}=R_1[1+\alpha (T_2-T_1)][/tex] ---------------------- (1)
in which R₁ is the resistance at temperature T₁
[tex]R_{T(2)[/tex] is the resistance at temperature T₂
Given that V= IR
R = [tex]\frac{V}{I}[/tex]
Therefore, the resistance at temperature 28°C is;
[tex]R_{28}= \frac{120V}{1.36A}[/tex]
= 88.24Ω
[tex]R_{T(2)[/tex] = [tex]\frac{120V}{1.23A}[/tex]
= 97.56Ω
From (1) above;
[tex]R_{T(2)}=R_1[1+\alpha (T_2-T_1)][/tex]
97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]
[tex]\frac{97.56}{88.24}= 1+(4.5*10^{-4})(T-28^0C)[/tex]
1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)
0.1056 = 4.5×10⁻⁴(T₂-28°C)
[tex]\frac{0.1056}{4.5*10^{-4}}= T-28^0C[/tex]
T - 28° C = 234.67
T = 234.67 + 28° C
T = 262.67 ° C
(b)
What is the power dissipated in the heating element initially and when the current reaches 1.23 A
The power dissipated in the heating element initially can be calculated as:
P = I²₁R₂₈
P = (1.36A)²(88.24Ω)
P = 163.209 W
P ≅ 163.21 W
The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:
[tex]P= I^2_2R_{T^0C[/tex]
P = (1.23)²(97.56Ω)
P = 147.598524
P ≅ 147.60 W
